Question

if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius

Answer 1

The mass of ore required is 21 700 t.

r = 750 cm

V =
`(4)/(3) \pi r^(3)` =
`(4)/(3) \pi (750 cm)^(3)` = 1.767 × 10⁹ cm³

The density of lead is 11.34 g/cm³.

So mass of lead sphere = 1.767 × 10⁹ cm³ ×
`(11.34 g)/(1 cm^(3) )` = 2.004 ×10¹⁰ g

2.004 ×10¹⁰ g ×
`(1 kg)/(1000 g)` = 2.004 × 10⁷ kg

2.004 × 10⁷ kg ×
`(1 t)/(1000 kg)` = 2.004 × 10⁴ t

92.5% efficiency means 92.5 t Pb per 100 t of ore.

Mass of ore = 2.004 × 10⁴ t Pb ×
`(100 tore)/(92.5 t Pb)` = 2.17 × 10⁴ t ore = 21 700 t ore